Question

The first couple of minutes of the launch of a space shuttle can be described very roughly as follows: The initial mass is \(2 \times 10^{6} \mathrm{kg}\), the final mass (after 2 minutes) is about \(1 \times 10^{6} \mathrm{kg}\), the average exhaust speed \(v_{\mathrm{ex}}\) is about \(3000 \mathrm{m} / \mathrm{s},\) and the initial velocity is, of course, zero. If all this were taking place in outer space, with negligible gravity, what would be the shuttle's speed at the end of this stage? What is the thrust during the same period and how does it compare with the initial total weight of the shuttle (on earth)?

Short answer

Speed is 2079 m/s, thrust is 25,000,000 N, and it's 1.27 times the initial weight.

Step-by-step solution

Understand the Rocket Equation

The rocket equation, or Tsiolkovsky's rocket equation, is used to estimate the final velocity (\(v_f\)) of a spacecraft given the initial and final masses, alongside the exhaust velocity. The equation is given by:\[v_f = v_{ ext{ex}} imes ext{ln}rac{m_i}{m_f}\]where \(v_{ ext{ex}}\) is the exhaust speed, \(m_i\) is the initial mass, and \(m_f\) is the final mass.

Calculate the Shuttle's Speed

Apply the values from the problem into the rocket equation:\[v_f = 3000 imes ext{ln}rac{2 imes 10^{6}}{1 imes 10^{6}}\]Calculate the natural logarithm:\[ ext{ln}rac{2 imes 10^{6}}{1 imes 10^{6}} = ext{ln}(2) \]This is approximately 0.693. Therefore:\[v_f = 3000 imes 0.693 = 2079 ext{ m/s}\]The shuttle's speed at the end of this stage is approximately 2079 m/s.

Determine the Thrust

Thrust can be calculated using the equation:\[ ext{Thrust} = rac{dm}{dt} imes v_{ ext{ex}}\]The rate of mass expulsion \(\frac{dm}{dt}\) can be found by dividing the total change in mass by the time of expulsion:\[rac{dm}{dt} = rac{(2 imes 10^{6} - 1 imes 10^{6})}{120 ext{ seconds}} = rac{1 imes 10^{6}}{120}\]Thus the thrust is:\[ ext{Thrust} = rac{1 imes 10^{6}}{120} imes 3000 = 25,000,000 ext{ N}\]The thrust during this period is 25,000,000 N.

Compare Thrust with Initial Weight

Calculate the initial weight of the shuttle by multiplying the initial mass by Earth's gravitational acceleration (\(g = 9.81 ext{ m/s}^2\)):\[ ext{Weight}_{ ext{initial}} = 2 imes 10^{6} imes 9.81 = 19,620,000 ext{ N}\]Now compare this with the thrust:\[rac{ ext{Thrust}}{ ext{Weight}_{ ext{initial}}} = rac{25,000,000}{19,620,000} \]This gives approximately 1.27, indicating the thrust is about 1.27 times the initial weight of the shuttle on Earth.

Key concepts

Thrust calculation

Calculating the thrust of a rocket engine is important to understand its capability to lift off from the Earth or maneuver in space. Thrust is generated by expelling mass at high speeds in the opposite direction of desired movement, following Newton's third law of motion: for every action, there is an equal and opposite reaction.

The formula for thrust (\[\text{Thrust} = \frac{dm}{dt} \times v_{\text{ex}}\]), involves two key components:

• Mass expulsion rate (\(\frac{dm}{dt}\)): This is the rate at which the rocket expels its propellant mass during flight.
• Exhaust velocity (\(v_{\text{ex}}\)): The speed at which propellant leaves the rocket engine.

For example, if a rocket expels gas at a rate of 1 million kilograms over 120 seconds with an exhaust velocity of 3000 m/s, the calculated thrust would be 25,000,000 N. This high thrust allows the rocket to overcome Earth's gravitational pull and move upward.

Space shuttle launch

The launch of a space shuttle is a carefully orchestrated event that involves the synchronization of several complex systems. A key objective during launch is to overcome Earth's gravity and achieve orbital velocity.

During the initial phase of launch, a shuttle's powerful engines are ignited to produce sufficient thrust to balance and eventually surpass the gravitational pull, allowing ascent into space. The structural integrity, aerodynamic design, and precision timing all play a critical role to ensure safety and efficiency.

• Shuttles start at massive initial masses, often exceeding millions of kilograms due to the fuel and structures required for the journey.
• As launch proceeds, fuel is consumed, reducing the total mass and thus increasing acceleration.
• The contrast between the initial total weight and provided thrust is a crucial factor that determines a successful launch.

The calculation of thrust, as seen with an initial mass of 2 million kg producing thrust well above its weight, demonstrates the capability to initiate liftoff safely.

Exhaust velocity

Exhaust velocity refers to the speed at which gases are expelled from a rocket engine. It is a fundamental aspect that dictates the efficiency and performance of a rocket.

Higher exhaust velocities generally translate to more efficient engines, as they can achieve greater thrust for a given rate of mass expulsion. In the context of the rocket equation, exhaust velocity (\(v_{\text{ex}}\)) is directly proportional to the overall change in velocity (\(v_f\)) of the spacecraft.

• An average exhaust velocity of 3000 m/s, like in the example, allows the rocket to gain significant speed within a short timeframe.
• By optimizing the exhaust velocity, engineers can enhance rocket propulsion while managing fuel consumption effectively.

Exhaust velocity, therefore, plays a vital role in determining how quickly a shuttle can ascend and how much speed it can ultimately attain, impacting the mission's success directly.

Mass expulsion rate

The mass expulsion rate is the rate at which a rocket expels its fuel or propellant. This factor is essential for calculating thrust, and it influences the rocket's velocity as described in the rocket equation.

In our context, a mass expulsion rate is calculated by dividing the change in mass (\(\Delta m\)) over a specific time period (\(\Delta t\)):\[\frac{dm}{dt} = \frac{\Delta m}{\Delta t}\]

• For example, if the shuttle reduces its mass by 1 million kg in 120 seconds, the mass expulsion rate is approximately 8333 kg/s.
• This measure helps determine how efficiently a rocket uses its propellant, which is crucial for achieving desired speeds and thrust effectively.

Understanding the mass expulsion rate is critical for optimizing flight performance and ensuring that the shuttle can reach its intended destination while maintaining fuel efficiency.